Question 11 Mark
Answer
View full question & answer→- 130°
Solution:
$\angle2 = 180^\circ-\angle1$
$\angle2 = 180^\circ-50^\circ=130^\circ$
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M.C.Q
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Question 21 Mark
Answer
View full question & answer→- 80º
Solution:
Since AOB is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow(3\text{x}-10)+50+(\text{x}+20)=180^\circ$
$\Rightarrow4\text{x}+60=180$
$\Rightarrow4\text{x}=120$
$\Rightarrow\text{x}=30$
So, $\angle\text{AOC}=3\text{x}-10=3(30)-10=80^\circ$
Question 31 Mark
View full question & answer→
Question 41 Mark
In the adjoining figure, $AOB$ is a straight line. If $x : y : z = 4 : 5 : 6,$ then $y =$?
Answer
View full question & answer→Let $\angle\text{AOC}=\text{x}^\circ=(4\text{a})^\circ,\angle\text{COD}=\text{y}^\circ=(5\text{a})^\circ$ and $\angle\text{BOD}=\text{z}^\circ=(6\text{a})^\circ$
Then, we have
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$ $[$Since $AOB$ is a straight line$]$
$\Rightarrow 4a + 5a + 6a= 180^\circ$
$\Rightarrow 15a = 180^\circ$
$\Rightarrow a = 12^\circ$
$\therefore y = 5 \times a = 5 \times 12^\circ = 60^\circ .$
Then, we have
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$ $[$Since $AOB$ is a straight line$]$
$\Rightarrow 4a + 5a + 6a= 180^\circ$
$\Rightarrow 15a = 180^\circ$
$\Rightarrow a = 12^\circ$
$\therefore y = 5 \times a = 5 \times 12^\circ = 60^\circ .$
Question 51 Mark
In figure, if $l_1|| l_2,$ what is $x + y$ in terms of $w$ and $z$?
Answer
Let angle supplement of $w^\circ$ be $a^\circ $.
$\Rightarrow a^\circ = 180^\circ - w^\circ$
Now,
$a^\circ = x^\circ [$Alternate opposite angles$]$
$\Rightarrow x^\circ = 180 - w^\circ ...(1)$
Now,
$y^\circ = z^\circ ...(2) [$Alternate angles$]$
Adding $(1)$ and $(2),$ we get
$x^\circ + y^\circ = 180^\circ - w^\circ + z^\circ$
View full question & answer→Let angle supplement of $w^\circ$ be $a^\circ $.
$\Rightarrow a^\circ = 180^\circ - w^\circ$
Now,
$a^\circ = x^\circ [$Alternate opposite angles$]$
$\Rightarrow x^\circ = 180 - w^\circ ...(1)$
Now,
$y^\circ = z^\circ ...(2) [$Alternate angles$]$
Adding $(1)$ and $(2),$ we get
$x^\circ + y^\circ = 180^\circ - w^\circ + z^\circ$
Question 61 Mark
Write the correct answer in the following:
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is.
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is.
Answer
View full question & answer→- An equilateral triangle.
Solution:
Let the angles of $\Delta\text{ABC}\ \text{be}\ \angle\text{A},\angle\text{B}\ \text{and}\ \angle\text{C}$
Given that $\angle\text{A}=\angle\text{B}+\angle\text{C}.....(1)$
But, in any $\Delta\text{ABC},\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ....(2)$[Angles sum property of triangle]
From equation (1) and (2), we get
$\angle\text{A}+\angle\text{A}=180^\circ\Rightarrow2\angle\text{A}=180^\circ\Rightarrow\angle\text{A}\frac{180^\circ}{2}=90^\circ$
$\Rightarrow\angle\text{A}=90^\circ$
Hence, the triangle is a right triangle and option (d) is correct.
Question 71 Mark
Answer
View full question & answer→- Each pair of co-interior angles is complementary.
Solution:
Co-interior angles are supplementary, not complementary.
Question 81 Mark
Answer
View full question & answer→- 80º
Solution:
Since AOB is a straight line,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow4\text{x}+5\text{x}=180^\circ$
$\Rightarrow9\text{x}=180^\circ$
$\Rightarrow\text{x}=206^\circ$
So, $\angle\text{AOC}=4\text{x}=4(20)=80^\circ$
Question 91 Mark
The sum of all the angles of a quadrilateral is:
Answer
View full question & answer→- 360º
Solution:
Sum of the angles of a polygon = (n - 2) × 180
Quadrilateral has 4 sides,
So sum of interior angles = (4 - 2) × 180º = 360
Question 101 Mark
How many triangles can be drawn having angles as 45º, 60º and 85º?
Answer
View full question & answer→- None.
Solution:
If we add up the three given angles we get
45º + 60º + 85º = 190º
But as we know that the angles of any triangle only add-up to 180º.
Hence the above-given triangle is not possible.
Question 111 Mark
View full question & answer→
Question 121 Mark
The sides of a quadrilateral are extended in order to form 4 exterior angles. The sum of these exterior angles is:
Question 131 Mark
Answer
View full question & answer→- 125º, 125º, 35º
Solution:
x + 55 = 180º (Sum of supplementary angles or co-interior angles)
x = 125º
x = y = 125º (Corresponding angles)
$\text{z}+\angle\text{EAB}=\text{y} $ (Exterior angle property)
z = 125º - 90º = 35º.
Question 141 Mark
Answer
View full question & answer→- 360º
Solution:
We know that the exteior angle is sum of opposite interior angle
$\angle\text{ACD}=\angle1+\angle2\text{ (i)}$
$\angle\text{BAE}=\angle3+\angle2\text{ (ii)}$
$\angle\text{CBF}=\angle1+\angle3\text{ (iii)}$
ADD Equation (i), (ii) and (iii)
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times(\angle1+\angle2+\angle3)$
$\angle1+\angle2+\angle3=180^\circ\text{(ASP)}$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times180^\circ$
$\angle\text{ACD} +$ We know that the exteior angle is sum of opposite interior angle
$\angle\text{ACD}=\angle1+\angle2\text{ (i)}$
$\angle\text{BAE}=\angle3+\angle2\text{ (ii)}$
$\angle\text{CBF}=\angle1+\angle3\text{ (iii)}$
ADD Equation (1i), (ii) and (iii)
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times(\angle1+\angle2+\angle3)$
$\angle1+\angle2+\angle3=180^\circ\text{(ASP)}$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times180^\circ$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=360^\circ.$
Question 151 Mark
The number of lines that can pass through a given point is:
Question 161 Mark
In Fig., if $l \ || \ m,$ then $x =$
Answer
View full question & answer→Given that,
$l \ || \ m$ and $n$ cuts them
Let,
$\angle1=65^\circ$
$\angle2=\text{x}$
$\angle3=40^\circ$
$\angle1+\angle4=65^\circ$ $($Alternate angle$) (i)$
$\angle3+\angle4+\angle5=180^\circ$ $($Angle sum property$)$
$40^\circ+165^\circ+\angle5=180^\circ$
$\angle5=75^\circ$
Now,
$\angle2+\angle5=180^\circ$ $($Linear pair$)$
$x + 75^\circ = 180^\circ$
$x = 105^\circ .$
$l \ || \ m$ and $n$ cuts them
Let,
$\angle1=65^\circ$
$\angle2=\text{x}$
$\angle3=40^\circ$
$\angle1+\angle4=65^\circ$ $($Alternate angle$) (i)$
$\angle3+\angle4+\angle5=180^\circ$ $($Angle sum property$)$
$40^\circ+165^\circ+\angle5=180^\circ$
$\angle5=75^\circ$
Now,
$\angle2+\angle5=180^\circ$ $($Linear pair$)$
$x + 75^\circ = 180^\circ$
$x = 105^\circ .$
Question 171 Mark
Answer
View full question & answer→- 130º
Solution:
$\angle2=180^\circ-\angle1$
$\angle2=180^\circ-50^\circ=130^\circ.$
Question 181 Mark
Answer
View full question & answer→- 80º
Solution:
Given that,
AB || HF and CD cuts them
$\angle\text{HFC}=\angle\text{FDA}$ (Corresponding angle)
$\angle\text{FDA}=28^\circ$
$\angle\text{FDA}+\angle\text{FDE}+\angle\text{EDB}=180^\circ $(Linear pair)
$28^\circ+\angle\text{FDE}+72^\circ=180^\circ$
$\angle\text{FDE}=80^\circ.$
Question 191 Mark
View full question & answer→
Question 201 Mark
Answer
View full question & answer→- 10º
Solution:
2x - 20º + x + 100º = 110º (Exterior angle is equal to sum of its interior opposite angles)
3x = 110º - 100º + 20º
x = 10º.
Question 211 Mark
In Fig. the value of $y$ is:
Answer
View full question & answer→$3x + y + 2x = 180^\circ ($Linear pair$)$
$5x + y = 180^\circ (i)$
From figure,
$y = x ($Vertically opposite angles$)$
Using it in $(i),$ we get
$5x + x = 180^\circ$
$6x = 180^\circ$
$x = 30^\circ$
Thus,
$y = x = 30^\circ .$
$5x + y = 180^\circ (i)$
From figure,
$y = x ($Vertically opposite angles$)$
Using it in $(i),$ we get
$5x + x = 180^\circ$
$6x = 180^\circ$
$x = 30^\circ$
Thus,
$y = x = 30^\circ .$
Question 221 Mark
Answer
View full question & answer→- 20º
Solution:
Since, POQ is a line segment.
$\therefore\angle\text{POQ}=180^\circ$
$\Rightarrow\angle\text{POA}+\angle\text{AOB}+\angle\text{BOQ}=180^\circ$
$\Rightarrow40^\circ+4\text{x}+3\text{x}=180^\circ$
[Putting $\angle\text{POA}=40^\circ,\angle\text{AOB}=4\text{x}$ and $\angle\text{BOQ}=3\text{x}$]
$\Rightarrow7\text{x}=180^\circ-40^\circ$
$\Rightarrow7\text{x}=140^\circ$
$\text{x}=\frac{140^\circ}{7}$
$\text{x}=20^\circ.$
Question 231 Mark
An angle is one fifth of its supplement. The measure of the angle is:
Answer
View full question & answer→- 30º
Solution:
Let the measure of the angle be xº.
So, its supplement = (180º - x)
According to the given condition,
$\text{x}=\frac{1}{5}(180^\circ-\text{x)}$
$\Rightarrow5\text{x}=180-\text{x}$
$\Rightarrow6\text{x}=180$
$\Rightarrow\text{x}=30^\circ$
Question 241 Mark
Two straight lines AB and CD cut each other at O. If $\angle\text{BOD}=63^\circ,$ then $\angle\text{BOC}=$
View full question & answer→
Question 251 Mark
Given two distinct points P and Q in the interior of $\angle\text{ABC},$ then $\overrightarrow{\text{AB}}$ will be:
Answer
View full question & answer→- On the $\angle\text{ABC}$
Solution:
$\overrightarrow{\text{AB}}$ is a line which from $\angle\text{ABC}$ and it is the part of $\triangle\text{ABC}.$
Question 261 Mark
Given $\angle\text{POR}=3\text{x}$ and $\angle\text{QOR}=2\text{x}+10^\circ.$If POQ is a straight line, then the value of x is:
View full question & answer→
Question 271 Mark
If $\angle\text{A}=4\angle\text{B} = 6\angle\text{C},$ then A : B : C?
Answer
View full question & answer→- 12 : 3 : 2
Solution:
Let A be x
$\text{B}=\frac{1}{4}\text{x}$
$\text{C}=\frac{1}{6}\text{x}$
A : B : C
$\text{x}=\frac{1}{4}\text{x}:\frac{1}{6}\text{x}$
LCM of 4 and 6 is 12
12 : 3 : 2.
Question 281 Mark
Answer
View full question & answer→- 120º
Solution:
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{BCA}=180^\circ$ (Angle sum property)
$50^\circ+30^\circ+\angle\text{BCA}=180^\circ$
$\angle\text{BCA}=100^\circ$
In $\triangle\text{ECD}$
$\angle\text{ECD}+\angle\text{CDE}+\angle\text{CED}=180^\circ$ (Angle sum property)
$180^\circ-\angle\text{BCA}+40^\circ+\angle\text{CED}=180^\circ$
$\angle\text{CED}=100^\circ-40^\circ=60^\circ$
$\angle\text{CED}+\angle\text{AED}=180^\circ$ (Linear Pair)
$\angle\text{AED}=180^\circ-60^\circ=120^\circ.$
Question 291 Mark
Answer
View full question & answer→- 80º
Solution:
$\angle\text{A}=20^\circ$ (Vertical opp Angle)
In $\angle\text{ABC}$
The exterior angle so formed is equal to the sum of the two interior opposite angles.
$\angle\text{A}+\angle\text{ABC}=100^\circ$
$20^\circ+\angle\text{ABC}=100$
$\angle\text{ABC}=100^\circ-20^\circ$
$\angle\text{ABC}=180^\circ.$
Question 301 Mark
View full question & answer→
Question 311 Mark
Angles of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is:
Answer
View full question & answer→- 40º
Solution:
Given, the ratio of angles of a triangle is 2 : 4 : 3.
Let the angles of a triangle be $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$
$\angle\text{A}=2\text{x}, \angle\text{B}=4\text{x}$
$\angle\text{C}=3\text{x},\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
[sum of all the angles of a triangle is 180º]
$2\text{x}+4\text{x}+3\text{x}=180^\circ$
$9\text{x}=180^\circ$
$\text{x}=\frac{180^\circ}{9}=20^\circ$
$\angle\text{A}=2\text{x}=2\times20^\circ=40^\circ$
$\angle\text{B}=4\text{x}=2\times20^\circ=80^\circ$
$\angle\text{C}=3\text{x}=3\times20^\circ=60^\circ$
Hence, the smallest angle of a triangle is 40º.
Question 321 Mark
Answer
View full question & answer→- 18º
Solution:
In the given figure, we have xº, yº and zº forming a linear pair, therefore these must be supplementary.
That is,
x + y + z = 180º (1)
Also,
$\frac{\text{y}}{\text{x}}=5$
y = 5x (2)
And
$\frac{\text{z}}{\text{x}}=4$
z = 4x (3)
Substituting (ii) and (iii) in (i), we get:
x + 5x + 4x = 180º
10x = 180º
$\text{x}=\frac{180^\circ}{10}$
x = 18º.
Question 331 Mark
Answer
View full question & answer→- 25º
Solution:
Given that,
PQ || RS
$\angle\text{AEF}=95^\circ$
$\angle\text{BHS}=110^\circ$
$\angle\text{ABC}=\text{x}^\circ$
$\angle\text{AEF}=\angle\text{AGH}=95^\circ$ (Corresponding angles)
$\angle\text{AGH}+\angle\text{HGB}=180^\circ$ (Linear pair)
$95^\circ+\angle\text{HGB}=180^\circ$
$\angle\text{HGB}=85^\circ$
$\angle\text{BHS}+\angle\text{BHG}=180^\circ $(Linear pair)
$110^\circ+\angle\text{BHG}=180^\circ$
$\angle\text{BHG}=70^\circ$
In $\triangle\text{BHG},$
$\angle\text{BHG}+\angle\text{HGB}+\angle\text{GBH}=180^\circ$
$70^\circ+85^\circ+\angle\text{GBH}=180^\circ$
$\angle\text{GBH}=25^\circ$
Thus,
$\angle\text{ABC}=\angle\text{GBH}=25^\circ.$
Question 341 Mark
Answer
View full question & answer→- 75º
Solution:
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{CAB}=180^\circ$ (Angle sum property)
$\angle\text{BCA}=180^\circ-95^\circ-40^\circ=45^\circ$
$\angle\text{BCA}=\angle\text{ECD}=45^\circ$ (Vertically opposite angle)
In $\triangle\text{ECD}$
$\angle\text{ECD}+\angle\text{DEC}+\angle\text{CDE}=180^\circ$ (Angle sum property)
$\angle\text{ECD}=180^\circ-45^\circ-60^\circ=75^\circ.$
Question 351 Mark
Answer
View full question & answer→- 30º
Solution:
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$\angle\text{ABC}=180^\circ-90^\circ-40^\circ$
$\angle\text{ABC}=50^\circ$
In $\angle\text{BED}$
$\angle\text{BED}+\angle\text{EBD}+\angle\text{BDE}=180^\circ$ (Angle sum property)
$\angle\text{BDE}=180^\circ-50^\circ-100^\circ$
$\angle\text{BDE}=30^\circ.$
Question 361 Mark
Answer
View full question & answer→- 180º
Solution:
$\angle1=\angle5$ (Corresponding angle)
$\angle7=\angle5$ (Vertically Opposite angle)
$\angle4+\angle1=180^\circ$ (Linear Pair)
$\angle4+\angle7=180^\circ$ (From above equations).
Question 371 Mark
Answer
View full question & answer→- 60º
Solution:
The ratio of the angles is given to be 4 : 5 : 6.
So, let the measure of the angles be 4m, 5m and 6m.
Since AOB is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow4\text{m}+5\text{m}+6\text{m}=180^\circ$
$\Rightarrow15\text{m}=180$
$\Rightarrow\text{m}=12$
So, y = 5m = 5(12) = 60º.
Question 381 Mark
An exterior angle of a triangle is $80^\circ$ and two interior opposite angles are equal. Measure of each of these angles is:
Answer
View full question & answer→- 40º
Solution:
We know that the exterior angle so formed is equal to the sum of the two interior opposite angles.
Let the two interior opposite angles be x.
So, x + x = 80º
2x = 80º
x = 40º.
Question 391 Mark
Answer
View full question & answer→- 100º
Solution:
Given that,
AB || CD
Produce P to Q so that PQ || AB || CD
$\angle\text{BAP}+\angle\text{APQ}=180^\circ$ (Interior angle)
$132^\circ+\angle\text{APQ}=180^\circ$
$\angle\text{APQ}=48^\circ\text{(i)}$
$\angle\text{APC}=\angle\text{APQ}+\angle\text{QPC}$
$148^\circ=48^\circ+\angle\text{QPC}$ [From (i)]
$\angle\text{QPC}=100^\circ$
$\angle\text{QPC}+\angle\text{PCD}=180^\circ$ (Interior angles)
$100^\circ+\angle\text{PCD}=180^\circ$
$\angle\text{PCD}=80^\circ$
$\angle\text{PCD}+\text{x}=180^\circ$(Linear pair)
$80^\circ+\text{x}=180^\circ$
$\text{x}=100^\circ.$
Question 401 Mark
An exterior angle of a triangle is 110º and its two interior opposite angles are equal. Each of these equal angles is:
Answer
View full question & answer→- 55º
Solution:
Let the measure of each of the two equal interior opposite angles of the triangle be x.
In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.
$\therefore$ x + x = 110º
⇒ 2x = 110º
⇒ x = 55º
Thus, the measure of each of these equal angles is 55º.
Question 411 Mark
The number of triangles that can be drawn having angles as 50º, 60º and 70º are:
Answer
View full question & answer→- Infinite.
Solution:
As we know similar triangles can be drawn for any given triangle.
These similar triangles will have the same angles as the original triangle (ie., $\angle50^\circ,\angle60^\circ$and $\angle70^\circ$) and will be infinite in number.
Question 421 Mark
Answer
View full question & answer→- 30º
Solution:
In $\triangle\text{OAB},$ we have
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ (Angle sum property)
$\Rightarrow55^\circ+75^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=50^\circ$
$\Rightarrow\angle\text{COD}=\angle\text{AOB}=50^\circ$ (Vertivcally opposite angles)
In $\triangle\text{OCD},$ we have
$\angle\text{COD}+\angle\text{OCD}+\angle\text{ODC}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+100^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$
Question 431 Mark
If two angles are supplementary and the larger is 20º less then three times the smaller, then the angles are:
Answer
View full question & answer→- $130^\circ,50^\circ$
Solution:
Let the two supplementary angles be xº and 180º - xº
Let 180º - x = 3x 20º
4x = 200º
x = 50º
So the angles are 50º and 130º.
Question 441 Mark
Answer
View full question & answer→- 56º
Solution:
x + 51º = 107º (Alternate interior angles)
x = 107º - 51º = 56º.
Question 451 Mark
The sum of the angles of a triangle is:
Question 461 Mark
The measure of the Complementary angle of $63^\circ$ is:
Answer
View full question & answer→Sum of complementary angles is $90^\circ .$
If one angle is $63^\circ$
Then the other angle will be $90^\circ - 63^\circ = 27^\circ .$
If one angle is $63^\circ$
Then the other angle will be $90^\circ - 63^\circ = 27^\circ .$
Question 471 Mark
In Fig. if $l1 || l2,$ what is the value of $y$?
Answer
View full question & answer→Given that,
$l1 || l2$ and $l3$ is transversal
$\angle1=3\text{x} ($Vertically opposite angle$)$
$\text{y}=\angle1 ($Corresponding angle$)$
$y = 3x (i)$
$y + x = 180^\circ ($Linear pair$)$
$3x + x = 180^\circ [$From $(i)]$
$4x = 180^\circ$
$x = 45^\circ$
Therefore,
$y = 3x = 3 \times 45^\circ$
$= 135^\circ .$
$l1 || l2$ and $l3$ is transversal
$\angle1=3\text{x} ($Vertically opposite angle$)$
$\text{y}=\angle1 ($Corresponding angle$)$
$y = 3x (i)$
$y + x = 180^\circ ($Linear pair$)$
$3x + x = 180^\circ [$From $(i)]$
$4x = 180^\circ$
$x = 45^\circ$
Therefore,
$y = 3x = 3 \times 45^\circ$
$= 135^\circ .$
Question 481 Mark
AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of $\angle\text{FEB}.$ If $\angle\text{LEB}=35^\circ,$ then $\angle\text{CFQ}$ will be:
View full question & answer→
Question 491 Mark
Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is:
Answer
View full question & answer→- 36º
Solution:
Let x and (90º - x) be two complimentary angles
According to question,
2x = 3 (90º - x)
2x = 270º - 3x
x = 54º
The angles are:
54º and 90º - 54º = 36º
Thus, smallest angle is 36º.
Question 501 Mark
Write the correct answer in the following:
An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is,
An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is,
Answer
View full question & answer→- $52\frac{1}{2}^\circ$
Solution:
An exterior angle of triangle is 150°
Let each of to two interior opposites angles be x.
We know that exterior angle of a equal to the sum of two interior opposite angles.
$105^\circ=\text{x}+\text{x}\Rightarrow2\text{x}=105^\circ$
$\Rightarrow\text{x}=\frac{1}{2}\times105^\circ=52\frac{1}{2}^\circ$
So, each of equal angle angle is $52\frac{1}{2}^\circ$
Hence, (b) is the correct answer.